## Solving inequalities, need some confirmation

I have these three inequalities that I am supposed to solve, I think I came up with the right answer but I'm not even 100% sure it's in the correct format.

A. 6x^2 < 6+5x
my work:
6x^2-5x-6 < 0
solutions are then 3/2 and -2/3
so the answer I got is:
-2/3 < x < 3/2

B. x^2+8x > 0
my work:
soutions I got were 0, -8
-8 < x < 0

C. (x+2)(x^2-x+1) > 0
my work:
x+2 > 0 and x^2-x+1 > 0
solutions are then -.414 and 2.414
-4.14 < x < 2.414
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 Quote by ability B. x^2+8x > 0 my work: soutions I got were 0, -8 so my answer is: -8 < x < 0
$$y = x^2+8x$$

Opens up...the vertex is at (-4, -16)...you found the zeros at 0 and -8.
Graph this....Now find the parts of the graph that are BIGGER than 0...in other words, what parts of the graph are above the x-axis?

 C. (x+2)(x^2-x+1) > 0 my work: x+2 > 0 and x^2-x+1 > 0 solutions are then -.414 and 2.414 so my answer is: -4.14 < x < 2.414
This has only one zero....x = -2.
$$x^2 - x +1 = (x - 1/2)^2 + 3/4$$

Which is clearly always above the x-axis and so you have no real roots....

So to the left of ( -2, 0)...for example x = -3, what would your expression evaluate to? A positive or negative number?
 Recognitions: Gold Member Homework Help Science Advisor No, no no! I'll take B for you: $$x^{2}+8x>0$$ This can be rewritten as: $$x(x+8)>0$$ What you call "solutions", are the solutions to the equation $$x^{2}+8x=0$$ These values are important in determining the regions of x-values where the INEQUALITY holds, but they are by no means indicative of these regions in the manner you think. Let us go back to: $$x(x+8)>0$$ The left-hand side has two factors. The product of two numbers are positive if a) each factor is positive (that is, x>0 AND, x+8>0) OR b) each factor is negative (that is, x<0 AND x+8<0) Try now to identify the regions on the x-axis where the inequality holds!